# Olympiad Geometry Problem about Pentagon Inscribed in Circle

Someone gave me this question, and I seriously have no clue how to answer it. I've tried considering the centre, drawing diagrams, and searching up all the theorems in the internet but I can't find a way to solve it. Any help would be greatly appreciated.

There is a pentagon $$ABCDE$$ inscribed in a circle, with $$AB = 2, BC = 5, CD = 2, DE = 5$$ and $$AD = 8$$. What is the length of the line segment $$BE$$?

It feels like there isn't enough information. I don't even think we can find or use the radius of the circle. There must be a trick somewhere, right?

Let $$X$$ be the intersection of segments $$AD$$ and $$BE$$. From the repeating values in the lengths given, $$ABCD$$ and $$BCDE$$ are trapezoids, so $$BCDX$$ is a parallelogram, which means $$|DX|=5$$, $$|AX|=8-5=3$$, and $$|BX|=2$$. By the intersecting chords theorem, $$|XE|={15\over2}$$, whence $$|BE|={19\over2}$$.

• How do you confirm that $ABCD$ and $BCDE$ are trapezoids? More specifically, how does $BC=DE$ and $AB=CD$ imply that they are trapezoids? Commented Dec 19, 2023 at 8:58
• @Sahaj The points lie on a circle so $\angle OAB = \angle OBA = \angle OCD = \angle ODC$ and $\angle OBC = \angle OCB = \angle ODE = \angle OED$ and so $\angle ABC =\angle BCD =\angle CDE$ etc. Commented Dec 19, 2023 at 12:18
• Alternatively, $|AC|=|BD|$ (because they are corresponding sides of congruent triangles) and $\angle DAB = \angle ADC$ (they subtend equal arcs). A quadrilateral with equal diagonals and equal base angles is a trapezoid. Commented Dec 19, 2023 at 14:58

Consider the quadrilater $$ABCD$$. It is iscribed to the same circumference as $$ABCDE$$. If we call $$a=AB$$, $$b=BC$$,$$c=CD$$, $$d=AD$$ and $$s=\frac{1}{2}(a+b+c+d)$$ you can obtain the radius of such circle (see eg. this) $$R=\frac{1}{4} \sqrt{\frac{(ab+cd)(ac+bd)(ad+bc)}{(s-a)(s-b)(s-c)(s-d)}}$$ Let's call the center of the circumscribed circle $$O$$. By basic trigonometry you have that $$AE=2R \cos( \angle OAE)$$ You can consider the isosceles triangle $$ABO$$, $$BCO$$, $$CDO$$, $$DEO$$. As you know all the sides length, by using again basic trigonometry you can obtain the angles $$\angle ABC$$, $$\angle BCD$$, $$\angle CDE$$. Moreover you can obtain the angles $$\angle OAB$$, $$\angle OED$$.

As the sum of the angles of a pentagon is $$3 \pi$$ you obtain that $$\angle OAE=\frac{1}{2}\left(3\pi-\angle ABC-\angle BCD - \angle CDE-\angle OAB -\angle OED\right)$$ So we know the length of the side $$AE$$. Moreover we have $$\alpha:=\angle BAE=\angle OAE+ \angle OAB$$ Now consider the triangle $$ABE$$. By using the law of cosines we obtain $$BE=AB \cos(\alpha) \pm \sqrt{(AE)^2- (AB)^2 \sin^2(\alpha)}$$

Edit: I thought $$EA=8$$. Since my answer is voted positively, I will not delete it. Normaly, it should be deleted since it is irrelevant.

Let $$AC=BD=CE=x$$, the chords which make triangles with chords of length $$2$$ and $$5$$. Let $$BE=y$$.

By Ptolemy's Theorem on $$ABCE$$ we have $$2x+40=xy$$ and on $$BCDE$$ we have $$2y+25=x^2.$$ Hence $$2y^3+17y^2-92y-1500=0.$$

ABCD has sides 2,5,2,8. As these sides are for points on the circle, we know that arc length of AB = CD. Thus, angle BAD is over the same arc length as CDA, therefore these angles are the same. Let's mark them with α.

Now, it is easy to see that BCDE is also a trapezoid, except its sides are 5,2,5,X. Sides are given in the same order as in ABCD. BCD = 180-α so BED = α. Angles of these trapezoids are therefore the same. For the first trapezoid calculate difference of parallel line lengths to side line length and get ratio r = 3/2. The same ratio holds for the other trapezoid. Thus, X = 5*r+2 = 9.5.